If we look at the peak of the mains, 325V to 400V say, and assume CCM, the on time will be ~ 0.20
We need an L value to give us ~ 3A in the 65kHz period, V/L = di /dt, where V = Vin, di = 3A, dt = 3uS
ergo L = 325uH, and needs to handle 12A pk say
Given that Lmax = Nmax . Ae . Bmax / Ipk
and that for ETD59 Ae = 375mm^2 ( allowing for the gap ), we know L, Ae, Bmax ( 0.28T ) and Ipk ( 12A say )
therefore Lmax = Nmax . 8.75 E-6, for L = 325uH, N must be 37 Turns, => this will easily fit on an ETD59 bobbin with a given current of 8.5A rms
We need to calc the gap, AL ( nH/N^2 ) = Uo.Ur. Ae / Lgap, for the gap ( air ) Ur = 1, we know AL desired from...